Let $g$ be a polynomial function and let $g'$, its derivative, be defined as $g'(x)=-x^2(x+1)^2(x-1)^2 $. At how many points does the graph of $g$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
Explanation: We can find the relative extrema (i.e. minima and maxima) of $g$ by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $g'(x)=-x^2(x+1)^2(x-1)^2 $. $g'(x)=0$ for $x=-1,0,1$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-1$, $x=0$, and $x=1$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}2.5$ $\llap{-}1.5$ $\llap{-}0.5$ $0.5$ $1.5$ $2.5$ $(-\infty,\ \ \llap{-}1)$ $(\ \ \llap{-}1,0)$ $(0,1)$ $(1,\infty)$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $(-\infty,-1)$ $x=-2$ $g'(-2)=-36<0$ $g$ is decreasing $\searrow$ $(-1,0)$ $x=-\dfrac12$ $g'\left(-\dfrac12\right)=-\dfrac{9}{64}<0$ $g$ is decreasing $\searrow$ $(0,1)$ $x=\dfrac12$ $g'\left(\dfrac12\right)=-\dfrac{9}{64}<0$ $g$ is decreasing $\searrow$ $(1,\infty)$ $x=2$ $g'(2)=-36<0$ $g$ is decreasing $\searrow$ Now let's look at the critical points: $x$ Before After Verdict $-1$ $\searrow$ $\searrow$ Not an extremum $0$ $\searrow$ $\searrow$ Not an extremum $1$ $\searrow$ $\searrow$ Not an extremum Now we can see that $g$ has no relative minima.